Show already existing record from database when adding from gridview footer.

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Hi Friends,

My requirement is, i have textboxes in Gridview Footer, Inserting of records from footer textboxes is working fine for me.

My problem is.. Same record must not be inserted. which is already in database.

When i click add button in gridview footer, a popup message must be displayed. Record already exists.

Please can you show me how to do this, This is my code of Gridview footer in GV ROW COMMAND.

------------------------------------

This is my code.

C#

protected void GV_UserEntry_RowCommand(object sender, GridViewCommandEventArgs e)
    {
        if (e.CommandName.Equals("AddNew"))
        {
            TextBox TxtRollNo = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterRollNo);
            TextBox TxtStudentName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterStudentName");
            TextBox TxtStandard = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterStandard");
            TextBox TxtFathersName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterFathersName");
            TextBox TxtMothersName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterMothersName");
            TextBox TxtFamilyName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterFamilyName");
            TextBox TxtLocation = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterLocation");
         
            con.Open();

            SqlCommand cmd = new SqlCommand("INSERT into StudentDetailsTable (RollNo,StudentName,Standard,FathersName,MothersName,FamilyName,Location) VALUES ('" + TxtRollNo .Text + "','" + TxtStudentName .Text + "','" + TxtStandard .Text + "','" + TxtFathersName.Text + "','" + TxtMothersName.Text + "','" + TxtFamilyName.Text + "','" + TxtLocation .Text + "')", con);
            int result = cmd.ExecuteNonQuery();
            con.Close();


            if (result == 1)
            {
                Bind_StudentDetails();
                ScriptManager.RegisterStartupScript(this, this.GetType(), "RunCode", "javascript:alert('Student Details Added successfully');", true);
            }

        }

       
    }

Please help thanks.

Posted 4-Jun-13 21:11pm

Ubaid ur Rahman IT

Updated 4-Jun-13 23:09pm

Zafar Sultan

v2

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Rockstar_ 5-Jun-13 3:26am

Use if not exist keywords form sql and try to modify the sql querry, or before inserting the data first check with the columns which you are comparing, then based on the result , insert the data or give error message.

3 solutions

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Osman24 · Answer 1 · 2013-06-04T21:24:00

Solution 1

You can do this in two way. One way is add primary key to your table. If try to add same value, it will refuse you.

Second way is before insertion you can get a select query to check whether there is a record same as you are trying to add or not?

In first solution as I remembered you should catch an exception.

To show your message you can use javascript alert.

By the way do not use this

C#

SqlCommand cmd = new SqlCommand("INSERT into StudentDetailsTable (RollNo,StudentName,Standard,FathersName,MothersName,FamilyName,Location) VALUES ('" + TxtRollNo .Text + "','" + TxtStudentName .Text + "','" + TxtStandard .Text + "','" + TxtFathersName.Text + "','" + TxtMothersName.Text + "','" + TxtFamilyName.Text + "','" + TxtLocation .Text + "')", con);

Instead of this, you can use parameters. By using parameter you block all kind of sql injection.

Here is the explanation:

http://msdn.microsoft.com/tr-tr/library/system.data.sqlclient.sqlcommand.parameters.aspx[^]

Posted 4-Jun-13 21:24pm

Osman24

Comments

Ubaid ur Rahman IT 5-Jun-13 3:59am

This is Intranet Application. Please I need it here only.

Please can you help me how to do this.

Osman24 5-Jun-13 4:05am

Can you be more spesific about what you need. If you are, I can help as much as I can

Ubaid ur Rahman IT 5-Jun-13 4:07am

Osman brother, I need.... Same record must not be inserted. which is already in database. when add button click in gridview footer...

by the way above is my code.

Please help thanks.

Osman24 5-Jun-13 4:14am

I understand but here is the thing When you say same record what does it mean. It is all values are same or not?.

Ubaid ur Rahman IT 5-Jun-13 4:20am

ya, Rollno and StudentName must not be same.

suppose, if Same Rollno and StudentName already exists in database. Show error MSG. (Sorry, this record already exists, Please try another).

If it is new record, it must save in database otherwise already exist error.

Please help

thanks.

Osman24 5-Jun-13 4:23am

Ok forget the primary key. as I understand it is urgent. Use this

public int Count()
{
string query = "SELECT Count(*) FROM StudentDetailsTable where RollNo = '" + TxtRollNo .Text + "'"; ///add other columns
int Count = -1;

//Open Connection
if (this.OpenConnection() == true)
{
//Create Mysql Command
MySqlCommand cmd = new MySqlCommand(query, connection);

//ExecuteScalar will return one value
Count = int.Parse(cmd.ExecuteScalar()+"");

//close Connection
this.CloseConnection();

return Count;
}
else
{
return Count;
}
}

if this count bigger than 0 run javascript alert code else make insertion.

Ubaid ur Rahman IT 5-Jun-13 4:50am

actually, This is my code, it must check on add button click in Gridview footer.

This is my code.

Please can you add ur code here, in my code. am unable to understand.

protected void GV_UserEntry_RowCommand(object sender, GridViewCommandEventArgs e)
{
if (e.CommandName.Equals("AddNew"))
{
TextBox TxtRollNo = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterRollNo);
TextBox TxtStudentName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterStudentName");
TextBox TxtStandard = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterStandard");
TextBox TxtFathersName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterFathersName");
TextBox TxtMothersName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterMothersName");
TextBox TxtFamilyName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterFamilyName");
TextBox TxtLocation = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterLocation");

con.Open();

SqlCommand cmd = new SqlCommand("INSERT into StudentDetailsTable (RollNo,StudentName,Standard,FathersName,MothersName,FamilyName,Location) VALUES ('" + TxtRollNo .Text + "','" + TxtStudentName .Text + "','" + TxtStandard .Text + "','" + TxtFathersName.Text + "','" + TxtMothersName.Text + "','" + TxtFamilyName.Text + "','" + TxtLocation .Text + "')", con);
int result = cmd.ExecuteNonQuery();
con.Close();

if (result == 1)
{
Bind_StudentDetails();
ScriptManager.RegisterStartupScript(this, this.GetType(), "RunCode", "javascript:alert('Student Details Added successfully');", true);
}

}

}

AND ALSO I NEED A POPUP MESSAGE, FOR RECORD ALREADY EXISTS.

Please help Osman bhai, its urgent THANKS.

Osman24 5-Jun-13 5:03am

protected void GV_UserEntry_RowCommand(object sender, GridViewCommandEventArgs e)
{
if (e.CommandName.Equals("AddNew"))
{
TextBox TxtRollNo = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterRollNo);
TextBox TxtStudentName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterStudentName");
TextBox TxtStandard = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterStandard");
TextBox TxtFathersName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterFathersName");
TextBox TxtMothersName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterMothersName");
TextBox TxtFamilyName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterFamilyName");
TextBox TxtLocation = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterLocation");

con.Open();

SqlCommand cmd = new SqlCommand( "SELECT Count(*) FROM StudentDetailsTable where RollNo = '" + TxtRollNo.Text + "' StudentName = '"+TxtStudentName.Text + "'", con);
int count = int.Parse(cmd.ExecuteScalar()+"");

if(count == 0)
{

cmd = new SqlCommand("INSERT into StudentDetailsTable (RollNo,StudentName,Standard,FathersName,MothersName,FamilyName,Location) VALUES ('" + TxtRollNo .Text + "','" + TxtStudentName .Text + "','" + TxtStandard .Text + "','" + TxtFathersName.Text + "','" + TxtMothersName.Text + "','" + TxtFamilyName.Text + "','" + TxtLocation .Text + "')", con);
int result = cmd.ExecuteNonQuery();
if(result == 1)
{
Bind_StudentDetails();
ScriptManager.RegisterStartupScript(this, this.GetType(), "RunCode", "javascript:alert('Student Details Added successfully');", true);
}

}
else
{
ScriptManager.RegisterStartupScript(this, this.GetType(), "RunCode", "javascript:alert('Record already exists');", true);
}

con.Close();

}

}

try this I have not compiled. I think this will work also I jave not check your alert code. Are you sure ScriptManager.RegisterStartupScript(this, this.GetType(), "RunCode", "javascript:alert('Student Details Added successfully');", true); this code is running

Ubaid ur Rahman IT 5-Jun-13 5:33am

ya its working but problem in scriptmanager popup(record already exists) this is not popup ing, and student details added is popup ing.

Please help,

Thanks.

Osman24 5-Jun-13 6:35am

I did not understand the problem. What I understand is It does not insert same record but not show popup but when a new record inserted it shows popup. Is it true?

Ubaid ur Rahman IT 5-Jun-13 6:54am

ya, you are absolutely right.

Ubaid ur Rahman IT 5-Jun-13 6:55am

can you please give me a solution for this ?

Thanks in advance.

Osman24 5-Jun-13 7:10am

Ok use google chrome open console by using f12 and try to insert an existing record and look for an error and share with me. I'm waiting for you

Hemant761 · Answer 2 · 2013-06-05T03:42:00

It is much easier.

See you can do like this.

as you are going for insert data into Database before that you check that address is already exist by making database call.

Means create another stored procedure that will go and check that same details already exists if yes then call ResgisterStartupScript that will say " Record already exist please enter different record."

or if no then insert data into database and show message record added successfully.

Srinivas Karra · Answer 3 · 2013-08-08T02:29:00

protected void GV_UserEntry_RowCommand(object sender, GridViewCommandEventArgs e)
{
if (e.CommandName.Equals("AddNew"))
{
TextBox TxtRollNo = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterRollNo);
TextBox TxtStudentName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterStudentName");
TextBox TxtStandard = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterStandard");
TextBox TxtFathersName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterFathersName");
TextBox TxtMothersName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterMothersName");
TextBox TxtFamilyName = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterFamilyName");
TextBox TxtLocation = (TextBox)GV_UserEntry.FooterRow.FindControl("txtFooterLocation");

con.Open();

SqlCommand cmd = new SqlCommand("INSERT into StudentDetailsTable (RollNo,StudentName,Standard,FathersName,MothersName,FamilyName,Location) VALUES ('" + TxtRollNo .Text + "','" + TxtStudentName .Text + "','" + TxtStandard .Text + "','" + TxtFathersName.Text + "','" + TxtMothersName.Text + "','" + TxtFamilyName.Text + "','" + TxtLocation .Text + "')", con);
int result = cmd.ExecuteNonQuery();
con.Close();

if (result == 1)
{
Bind_StudentDetails();
ScriptManager.RegisterStartupScript(this, this.GetType(), "RunCode", "javascript:alert('Student Details Added successfully');", true);
}

}

}

Show already existing record from database when adding from gridview footer.

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